The Relational Data Model
Functional Dependencies
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Functional Dependencies
X -> A is an assertion about a relation R that
whenever two tuples of R agree on all the
attributes of X, then they must also agree on
the attribute A.
 Say “X -> A holds in R.”
 Notice convention: …,X, Y, Z represent sets of
attributes; A, B, C,… represent single attributes.
 Convention: no set formers in sets of attributes,
just ABC, rather than {A,B,C }.
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Example
 Drinkers(name, addr, beersLiked,
manf, favBeer).
 Reasonable FD’s to assert:
1. name -> addr
2. name -> favBeer
3. beersLiked -> manf
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Example Data
name
Janeway
Janeway
Spock
addr
Voyager
Voyager
Enterprise
Because name -> addr
beersLiked
Bud
WickedAle
Bud
manf
A.B.
Pete’s
A.B.
favBeer
WickedAle
WickedAle
Bud
Because name -> favBeer
Because beersLiked -> manf
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FD’s With Multiple Attributes
No need for FD’s with > 1 attribute on
right.
 But sometimes convenient to combine FD’s
as a shorthand.
 Example: name -> addr and name ->
favBeer become name -> addr favBeer
 > 1 attribute on left may be essential.
 Example: bar beer -> price
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Keys of Relations
 K is a key for relation R if:
1. Set K functionally determines all attributes of R
2. For no proper subset of K is (1) true.


If K satisfies (1), but perhaps not (2), then
K is a superkey.
Note E/R keys have no requirement for
minimality, as in (2) for relational keys.
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Example
Consider relation Drinkers(name, addr,
beersLiked, manf, favBeer).
 {name, beersLiked} is a superkey
because together these attributes
determine all the other attributes.
 name -> addr favBeer
 beersLiked -> manf
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Example, Cont.
{name, beersLiked} is a key because
neither {name} nor {beersLiked} is a
superkey.
 name doesn’t -> manf; beersLiked doesn’t
-> addr.
In this example, there are no other
keys, but lots of superkeys.
 Any superset of {name, beersLiked}.
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E/R and Relational Keys
Keys in E/R are properties of entities
Keys in relations are properties of tuples.
Usually, one tuple corresponds to one entity,
so the ideas are the same.
But --- in poor relational designs, one entity
can become several tuples, so E/R keys and
Relational keys are different.
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Example Data
name
Janeway
Janeway
Spock
addr
Voyager
Voyager
Enterprise
beersLiked
Bud
WickedAle
Bud
manf
A.B.
Pete’s
A.B.
favBeer
WickedAle
WickedAle
Bud
Relational key = name beersLiked
But in E/R, name is a key for Drinkers, and beersLiked is a key
for Beers.
Note: 2 tuples for Janeway entity and 2 tuples for Bud entity.
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Where Do Keys Come From?
1. We could simply assert a key K. Then
the only FD’s are K -> A for all
atributes A, and K turns out to be the
only key obtainable from the FD’s.
2. We could assert FD’s and deduce the
keys by systematic exploration.
 E/R gives us FD’s from entity-set keys
and many-one relationships.
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FD’s From “Physics”
While most FD’s come from E/R
keyness and many-one relationships,
some are really physical laws.
Example: “no two courses can meet in
the same room at the same time” tells
us: hour room -> course.
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Inferring FD’s: Motivation
In order to design relation schemas
well, we often need to tell what FD’s
hold in a relation.
We are given FD’s X1 -> A1, X2 -> A2,…,
Xn -> An , and we want to know
whether an FD Y -> B must hold in any
relation that satisfies the given FD’s.
 Example: If A -> B and B -> C hold, surely
A -> C holds, even if we don’t say so.
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Inference Test
To test if Y -> B, start assuming two tuples
agree in all attributes of Y.
Use the given FD’s to infer that these tuples
must also agree in certain other attributes.
If B is eventually found to be one of these
attributes, then Y -> B is true; otherwise, the
two tuples, with any forced equalities form a
two-tuple relation that proves Y -> B does
not follow from the given FD’s.
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Closure Test
An easier way to test is to compute the
closure of Y, denoted Y +.
Basis: Y + = Y.
Induction: Look for an FD’s left side X
that is a subset of the current Y +. If
the FD is X -> A, add A to Y +.
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X
Y+
A
new Y+
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Finding All Implied FD’s
Motivation: “normalization,” the process
where we break a relation schema into
two or more schemas.
Example: ABCD with FD’s AB ->C,
C ->D, and D ->A.
 Decompose into ABC, AD. What FD’s hold in
ABC ?
 Not only AB ->C, but also C ->A !
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Basic Idea
To know what FD’s hold in a projection,
we start with given FD’s and find all
FD’s that follow from given ones.
Then, restrict to those FD’s that involve
only attributes of the projected schema.
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Simple, Exponential Algorithm
1. For each set of attributes X, compute
X +.
2. Add X ->A for all A in X + - X.
3. However, drop XY ->A whenever we
discover X ->A.
 Because XY ->A follows from X ->A.
4. Finally, use only FD’s involving
projected attributes.
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A Few Tricks
Never need to compute the closure of
the empty set or of the set of all
attributes.
If we find X + = all attributes, don’t
bother computing the closure of any
supersets of X.
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Example
ABC with FD’s A ->B and B ->C.
Project onto AC.
 A +=ABC ; yields A ->B, A ->C.
• We do not need to compute AB
+
or AC +.
 B +=BC ; yields B ->C.
 C +=C ; yields nothing.
 BC +=BC ; yields nothing.
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Example, Continued
Resulting FD’s: A ->B, A ->C, and
B ->C.
Projection onto AC : A ->C.
 Only FD that involves a subset of {A,C }.
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A Geometric View of FD’s
Imagine the set of all instances of a
particular relation.
That is, all finite sets of tuples that
have the proper number of
components.
Each instance is a point in this space.
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Example: R(A,B)
{(1,2), (3,4)}
{}
{(5,1)}
{(1,2), (3,4), (1,3)}
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An FD is a Subset of Instances
 For each FD X -> A there is a subset
of all instances that satisfy the FD.
 We can represent an FD by a region in
the space.
 Trivial FD : an FD that is represented
by the entire space.
 Example: A -> A.
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Example: A -> B for R(A,B)
{(1,2), (3,4)}
A -> B
{}
{(5,1)}
{(1,2), (3,4), (1,3)}
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Representing Sets of FD’s
If each FD is a set of relation instances,
then a collection of FD’s corresponds to
the intersection of those sets.
 Intersection = all instances that satisfy all
of the FD’s.
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Example
Instances satisfying
A->B, B->C, and
CD->A
A->B
B->C
CD->A
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Implication of FD’s
If an FD Y -> B follows from FD’s
X1 -> A1,…, Xn -> An , then the region in
the space of instances for Y -> B must
include the intersection of the regions
for the FD’s Xi -> Ai .
 That is, every instance satisfying all the
FD’s Xi -> Ai surely satisfies Y -> B.
 But an instance could satisfy Y -> B, yet
not be in this intersection.
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Example
A->B A->C B->C
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